We start by choosing a Weyl quantization procedure \(\operatorname{\mathrm {Op}}\)  — as in §14 of [1]} — and write \(A = \operatorname{\mathrm {Op}}(a) + R_0\) , \(R_0\)  being a \(C^\infty \)  smoothing operator. Next, we introduce a positive parameter \(h>0\)  and a cutoff, writing \(a= a \chi + a (1-\chi )\) , so that \(a \chi \)  is supported in \(|\xi |\leqslant 1/h\) , and \(a(1-\chi )\)  in \(|\xi |> 1/2h\) .
By rescaling, we observe that \(\operatorname{\mathrm {Op}}(a(1-\chi ))\in \Psi ^0_{0,\, \mathrm {sc}}\)  of the form \(\operatorname{\mathrm {Op}}_h(b)\)  with
\(b(z,\xi ) = [a(1-\chi )](z, \xi /h).\) 
Using the strategy of [2]}, we then obtain that as long as \(0\leqslant \delta <1/2\)  and \(|t|\leqslant \delta |\log h|/ \lambda \) , we can find a symbol \(b_t\)  satisfying \(S_\delta ^0\)  estimates and such that
\(e^{tX} \operatorname{\mathrm {Op}}_h(b) e^{-tX} = \operatorname{\mathrm {Op}}_h(b_t) + \mathcal {O}_{\mathrm {smoothing}}(h^\infty ).\) 
We will not detail the arguments of [2]}, which are presented originally for operators in \(\mathbb {R}^n\) , with slightly different classes of symbols. For an exposition on surfaces, we refer to the appendix A of [4]}. The two main ingredients in the proof are the following

For the range of times allowed, \(b\circ \Phi _t\in S_\delta ^0\)  uniformly.

We have a product formula, for \(c\in S^m_\delta \) :
\([X, \operatorname{\mathrm {Op}}_h(c)] = \frac{h}{i} \operatorname{\mathrm {Op}}_h( \lbrace  p, c \rbrace ) + \mathcal {O}(h^{2-2\delta } \Psi ^{m-2}_{\delta ,\mathrm {sc}}).\) 

It is a feature of the proof that
\(b_t = b\circ \Phi _t + \mathcal {O}(h^{1-\delta } S^{-1}_\delta ),\) 
and \(b_t\)  shares the support of \(b\circ \Phi _t\) .
Now, \(\operatorname{\mathrm {Op}}(a\chi )\)  is not semi-classical, so that there is not much microlocal information we can find on \(e^{tX} \operatorname{\mathrm {Op}}(a \chi ) e^{-tX}\)  for \(t\)  large. The best we can do is thus estimating \(H^s\rightarrow H^r\)  norms. Letting
\(R_0^{\prime } = R_0+ \operatorname{\mathrm {Op}}(a\chi ),\) 
we get \(\Vert  R_0^{\prime } \Vert _{H^s\rightarrow H^r} = \mathcal {O}( h^{s-r} )\)  as \(h\rightarrow 0\)  when \(r\geqslant s\) . Then in the range \(|t|\leqslant \delta |\log h|/\lambda \) , we find
\(\Vert  e^{tX} R_0^{\prime } e^{-tX} \Vert _{H^{-N}\rightarrow H^N} &\leqslant C_N h^{-2N} \Vert  e^{-tX} \Vert _{H^N} \Vert  e^{tX} \Vert _{H^{-N}} \\&\leqslant C h^{-2N} e^{2N\lambda |t| + 2\mu |t|}\\&\leqslant C h^{-2N(1+\delta ) - 2\mu \delta /\lambda } \leqslant C h^{-3N - \mu /\lambda }.\) 

Hence by setting
\(A_t = \operatorname{\mathrm {Op}}_h(b_t)\quad \text{and} \quad R_t = e^{tX} R_0^{\prime } e^{-tX} + e^{tX} \operatorname{\mathrm {Op}}_h(b) e^{-tX} - A_t,\) 
the proof is complete.
